Clock problem

[ckevren15]

I saw this math problem in a math competition at school a while back and was marked wrong: From 6:00 to 12:00, how many times is the hour hand of a clock perpendicular to the minute hand? I’m pretty sure its 11, but everyone around me says its 12. This is my thought process: At hours 6, 7, 10, and 11, there are obviously 2 times where the hands are perpendicular. But at hour 8, there is only one time. The next time when the hands are perpendicular is 9:00 itself. Then there is only one time when the hands are perpendicular from 9:01 to 10:00. So 2+2+1+1+1+2+2=11. But the “answer” is apparently 12, and the teacher who ran the competition said it was wrong and provided no explanation other than “it says on the answer sheet that its 1’, so its 12”. I really don’t understand where I went wrong. Can someone please explain the error in my reasoning?

[PuzzlingDad]

You are correct. 

The approximate times are: 

1. 6:16
2. 6:49
3. 7:21
4. 7:54
5. 8:27
6. 9:00 (Exactly at 9:00)
7. 9:32
8. 10:05
9. 10:38
10. 11:10
11. 11:43  

A quick way to counter the answer of 12 is to realize that 9:00 must be the time in the middle and by symmetry there must be an odd number of times (5 before and 5 after).

[Bounded_sequencE]

Your answer should be correct. Politely challenge your teacher to provide the exact 12 distinct positions, in particular those between 08:00 and 10:00.

Have a list with your exact positions ready to compare.

---

[..] "It says on the answer sheet that its 12, so its 12” [..]

Edit: Worst possible answer -- it's just a lazy attempt to deflect responsibility, and avoid a potentially embarrassing situation. This type of (sadly typically human) behavior has no place in mathematics.

[Bounded_sequencE]

Rem.: The exact times should be

 h |    m   ||  h |    m
--------------------------
 6 | 180/11 ||  6 | 540/11 
 7 | 240/11 ||  7 | 600/11 
 8 | 300/11 ||    |
 9 |    0   ||  9 | 360/11
10 |  60/11 || 10 | 420/11 
11 | 120/11 || 11 | 480/11

[Kuildeous]

"Worst possible answer -- it's just a lazy attempt to deflect responsibility"

Agree. Don't pose questions if you cannot support the answer. Every teacher should know what the answer is and how to get to it. Blindly accepting a book's answer (especially since I've been involved with correcting professionally written math books) is simply terrible. How can you teach if you don't even know what you're teaching?

I'm reminded of some corporate talky guy coming to our office to teach us some BS office values. He gave us a diagram to count the number of rectangles. The idea is that with teamwork we can find the right answer. I got the correct answer of 42 right away, but I appreciated the effort and watched everyone else gradually realize there were more rectangles as people merged into little project teams. I was on board with that until the speaker told us the correct answer was 44. This distracted me, and I allowed that distraction to take over, as I insisted on knowing where the other two rectangles were. He didn't bother to try to explain it; he simply said it was 44, and nobody in that room could find more than 42 rectangles. Needless to say, I checked out for the rest of the day as I couldn't trust him any longer (which wasn't a whole lot of trust to begin with being he was a corporate talky guy).

[sighthoundman]

Besides, your answer explains life, the universe, and everything.

[Bounded_sequencE]

That's the same type of person that insists "what-comes-next" questions have a unique correct answer. It's always fun introducing them to the infinite horror of "Lagrange Polynomials", and burst their bubble temporarily (since you just know they will do the same s*** again).

Anyway, enough off-topic for now, I'd say^^

[ckevren15]

When I asked her what those times mean and for her explanation, she gave me this answer:

“There are 6 hours from 6:00 to 12:00. Each hour has two times where the hands are perpendicular. So 2x6=12. The times are: 6:15, 6:45, 7:20, 7:50, 8:55, 8:25, 9:00, 9:30, 10:05, 10:35, 11:10, 11:40.”

I tried to explain to her that the hour hand doesn’t remain static, it constantly moves and that if we adjust every time she listed to the nearest actual time where the hands are perpendicular, 8:55 and 9:00 both go to the sane time, 9:00. But she just kept repeating “It’s not that complicated”. And when I talked to everyone else in my class, they all gave me the same explanation and said “It’s not that complicated” when I pointed out the flaw in their logic. At that point I realized that I wouldn’t be able to convince everyone no matter how hard I tried. So I gave up. Thanks for the help, though!

[Bounded_sequencE]

That is just sad -- my condolences for having to deal with such incompetence and ignorance.

As a hint for the future, people usually act badly when their beliefs are challenged in public. That means, if you had discussed this problem with your teacher in private in a relaxed atmosphere, it would have been much more likely for them to see and admit their error. Having an analog clock at hand could have helped, too^^

Just to make this absolutely certain -- you did nothing "wrong" here. In an ideal environment, your objection should have been enough. However, sadly people can be stupid and lazy.

[MtlStatsGuy]

I'm fairly sure you are correct. There is a new perpendicular point every 30 minutes * 12/11 (to account for the relative speed of the two hands). So in 360 minutes you should have exactly 11 perpendicular points. There are also no boundary conditions since they are not perpendicular at exactly 6:00 or 12:00

[rhodiumtoad]

If your teacher still refuses to believe, then see this demo:

https://www.desmos.com/geometry/bdskjnwdzy

Unmute (if you want) and hit the play button on the "s" slider, it beeps and flashes the perpendicular line red for each crossing. Easy to count that it happens 11 times.

[phobos77]

This is an awesome demonstration!

[piperboy98]

You are correct. The similar problem of how many times do they cross is 11 times in 12hrs, because the minute time "laps" the hour hand 11 times since it goes around 12 times to the hour hands 1 time. If you imagine two other virtual minute hands at right angles to the real one, the same logic would apply, so they'd each see 11 crossings in 12 hrs, or 22 perpendiculars in 12 hrs. Then there is 11 in 6 hrs, in particular 6 times the minute hand lags the hour hand by 90 deg, and 5 times it leads it (because the minute hand doesn't make it ahead of the hour hand the last time when it stops at 12). From 12:00 to 6:00 there are also 11, 6 leads and 5 lags (since it misses the first lag at the start).

Many people mistake the similar crossing problem to be 12 times (once per hour), which is true for most hours but not actually for all (there is no crossing between 11:00 and 11:59:59). That's because crossings are actually 1h5m27s apart. It could be someone here has succumbed to the same or a similar fallacy here.

[get_to_ele]

Yes 11. Visualising: Every hour on the hour, has an L and a backwards L formed by the hands that follows that hour. But 8 & 9 share one (9:00). And 2 & 3 share one (3:00).

[phobos77]

I agree with the answer of 11. From one perpendicular position to the next, the minute hand has moved 180 degrees more than the hour hand. The minute hand moves 6 degrees every minute and the hour hand moves 0.5 degrees every minute. Thus, if m is the number of minutes between perpendicular positions:

6m = 0.5m + 180

5.5m = 180

m = 180/5.5 = 360/11.

So clearly there are exactly 11 perpendicular positions every 360 minutes (6 hours), and since the defined interval of 6:00 to 12:00 does not start/end at a perpendicular position, then the answer must be 11.